Reputation: 382
Reverse engineering "checksum" generation
I've been analyzing some 230V communication and trying to figure out the checksum algorithm of the protocol. From what I've learned the communication structure is the following:
- The first block is always the same, must be some sort of start structure (910hex)
- Then the nest 5 blocks (each of 4bit) are the actual data
- The last 8bit must be the checksum
I'm not sure if the checksum is only calculated over the data or the start structure as well (which is 910h in HEX)
I've attached some communication messages, each row is one message.
I've tried algorithms like CheckSum8 Xor, CheckSum8 Modulo 256 and CheckSum8 2s Complement; unfortunately without any luck.
| Start structure | Data | ChkSum |
00000100100010000 1010 1010 1010 1010 1011 10001100
00000100100010000 1010 1010 1010 1010 1001 01111100
00000100100010000 1010 1010 1010 1010 1101 01110010
00000100100010000 1010 1010 1010 1010 0101 01101110
00000100100010000 1010 1010 1010 1011 0101 01010110
00000100100010000 1010 1010 1010 1001 0101 00100110
00000100100010000 1010 1010 1010 1101 0101 11000110
00000100100010000 1010 1010 1010 0101 0101 11101000
00000100100010000 1010 1010 1011 0101 0101 10110100
00000100100010000 1010 1010 1001 0101 0101 00001100
00000100100010000 1010 1010 1101 0101 0101 10010010
00000100100010000 1010 1010 0101 0101 0101 01000000
00000100100010000 1010 1011 0101 0101 0101 00001010
00000100100010000 1010 1001 0101 0101 0101 10011110
00000100100010000 1010 1101 0101 0101 0101 01011000
00000100100010000 1010 0101 0101 0101 0101 00111010
00000100100010000 1011 0101 0101 0101 0101 11111110
00000100100010000 1001 0101 0101 0101 0101 10011000
00000100100010000 1101 0101 0101 0101 0101 01010100
00000100100010000 0101 0101 0101 0101 0101 00100010
UPDATE: I've made a much further progress now, and I'm currently no longer sure if it's a checksum or a CRC at all. I've generated and sent the data value from 0 to 34 and the system generated the following "checksums"
What I've learned so far:
- The high nibble (upper half-byte) changes only ever second time
- The low nibble is XORED with 1010, so bit 3 and 1 changes (not always)
- The high nibble is inverted every second time (not always)
- Bit 0 of the "checksum" is always 0
The HEX-Column at the end is the Checksum value in HEX
| Start structure | Data | ?ChkSum? | HEX |
00 00000100100010000 0000 0000 0000 0000 0000 1011 1000 B8
01 00000100100010000 0000 0000 0000 0000 0001 1011 0010 B2
02 00000100100010000 0000 0000 0000 0000 0010 0100 1000 48
03 00000100100010000 0000 0000 0000 0000 0011 0100 0010 42
04 00000100100010000 0000 0000 0000 0000 0100 1011 0110 B6
05 00000100100010000 0000 0000 0000 0000 0101 1011 1100 BC
06 00000100100010000 0000 0000 0000 0000 0110 0100 0110 46
07 00000100100010000 0000 0000 0000 0000 0111 0100 1100 4C
08 00000100100010000 0000 0000 0000 0000 1000 1010 0100 A4
09 00000100100010000 0000 0000 0000 0000 1001 1010 1110 AE
10 00000100100010000 0000 0000 0000 0000 1010 0101 0100 54
11 00000100100010000 0000 0000 0000 0000 1011 0101 1110 5E
12 00000100100010000 0000 0000 0000 0000 1100 1010 1010 AA
13 00000100100010000 0000 0000 0000 0000 1101 1010 0000 A0
14 00000100100010000 0000 0000 0000 0000 1110 0101 1010 5A
15 00000100100010000 0000 0000 0000 0000 1111 0101 0000 50
16 00000100100010000 0000 0000 0000 0001 0000 1000 0000 80
17 00000100100010000 0000 0000 0000 0001 0001 1000 1010 8A
18 00000100100010000 0000 0000 0000 0001 0010 0111 0000 70
19 00000100100010000 0000 0000 0000 0001 0011 0111 1010 7A
20 00000100100010000 0000 0000 0000 0001 0100 1000 1110 8E
21 00000100100010000 0000 0000 0000 0001 0101 1000 0100 84
22 00000100100010000 0000 0000 0000 0001 0110 0111 1110 7E
23 00000100100010000 0000 0000 0000 0001 0111 0111 0100 74
24 00000100100010000 0000 0000 0000 0001 1000 1001 1100 9C
25 00000100100010000 0000 0000 0000 0001 1001 1001 0110 96
26 00000100100010000 0000 0000 0000 0001 1010 0110 1100 6C
27 00000100100010000 0000 0000 0000 0001 1011 0110 0110 66
28 00000100100010000 0000 0000 0000 0001 1100 1001 0010 92
29 00000100100010000 0000 0000 0000 0001 1101 1001 1000 98
30 00000100100010000 0000 0000 0000 0001 1110 0110 0010 62
31 00000100100010000 0000 0000 0000 0001 1111 0110 1000 68
32 00000100100010000 0000 0000 0000 0010 0000 1100 1000 C8
33 00000100100010000 0000 0000 0000 0010 0001 1100 0010 C2
34 00000100100010000 0000 0000 0000 0010 0010 0011 1000 38
UPADTE 2: I've generated now the data, where only one bit of the data structure is 1
| Dec | Start structure | Data | ?ChkSum? | HEX |
00 00000100100010000 0000 0000 0000 0000 0000 1011 1000 B8
01 00000100100010000 0000 0000 0000 0000 0001 1011 0010 B2
02 00000100100010000 0000 0000 0000 0000 0010 0100 1000 48
04 00000100100010000 0000 0000 0000 0000 0100 1011 0110 B6
08 00000100100010000 0000 0000 0000 0000 1000 1010 0100 A4
16 00000100100010000 0000 0000 0000 0001 0000 1000 0000 80
32 00000100100010000 0000 0000 0000 0010 0000 1100 1000 C8
64 00000100100010000 0000 0000 0000 0100 0000 0101 1000 58
128 00000100100010000 0000 0000 0000 1000 0000 1001 0110 96
256 00000100100010000 0000 0000 0001 0000 0000 1110 0100 E4
512 00000100100010000 0000 0000 0010 0000 0000 0000 0000 00
1024 00000100100010000 0000 0000 0100 0000 0000 0010 0110 26
2048 00000100100010000 0000 0000 1000 0000 0000 0110 1010 6A
4096 00000100100010000 0000 0001 0000 0000 0000 1111 0010 F2
8192 00000100100010000 0000 0010 0000 0000 0000 0010 1100 2C
16384 00000100100010000 0000 0100 0000 0000 0000 0111 1110 7E
32768 00000100100010000 0000 1000 0000 0000 0000 1101 1010 DA
65536 00000100100010000 0001 0000 0000 0000 0000 0111 1100 7C
131072 00000100100010000 0010 0000 0000 0000 0000 1101 1110 DE
262144 00000100100010000 0100 0000 0000 0000 0000 0111 0100 74
524288 00000100100010000 1000 0000 0000 0000 0000 1100 1110 CE
Upvotes: 2
Views: 904
Answers (1)
Reputation: 1264
I think I've cracked this one. Here's how it works:
Start with your 20 input bits and an initial running result of
0b10111000(0x8B).There is a constant 8-bit mask for each of the 20 input bits. For each input bits with a
1value, you want to XOR the current running result with the 8-bit mask for that bit.
The solution in C:
const UINT8 bitMasks[20] = {
0b00001010,
0b11110000,
0b00001110,
0b00011100,
0b00111000,
0b01110000,
0b11100000,
0b00101110,
0b01011100,
0b10111000,
0b10011110,
0b11010010,
0b01001010,
0b10010100,
0b11000110,
0b01100010,
0b11000100,
0b01100110,
0b11011100,
0b01110110
};
UINT8 ComputeWeirdHash(UINT32 inBits)
{
UINT8 result = 0b10111000;
for (i=0; i<20; i++)
{
if (inBits & (1L << i))
result ^= bitMasks[i];
}
return result;
}
Now, to find each bitMask value, use an input where only one bit is 1. Then run it through your test system to find the 8-bit checksum. XOR that result with our initial running result value: 0b10111000, to find the bitMask value.
Upvotes: 2
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